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Math Help - issues with the algebra in a problem

  1. #1
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    issues with the algebra in a problem

    I have figured out how to solve this problem but I'm not getting the correct answer.

    A 1000 L tank loses water so that, after t days, the remaining volume is

    v(t) = 1000[1-(t/10)]^2 for 0<t<10.

    How rapidly is the water being lost when the tank is half full?

    I got the derivative fine. -200(1-(t/10))


    I know to solve v(t) = 500 then take the value of t and input it into v'(t)

    when solving v(t) = 500 I end up with the answer 10-5sqrt(2) or -5(sqrt(2) - 2)

    but to get the correct answer to the question it needs to be 5sqrt2

    the correct answer is 100sqrt(2) litres/day

    thank you for your help!
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  2. #2
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    Re: issues with the algebra in a problem

    Quote Originally Posted by Jonroberts74 View Post
    I have figured out how to solve this problem but I'm not getting the correct answer.

    A 1000 L tank loses water so that, after t days, the remaining volume is

    v(t) = 1000[1-(t/10)]^2 for 0<t<10.

    How rapidly is the water being lost when the tank is half full?

    I got the derivative fine. -200(1-(t/10))


    I know to solve v(t) = 500 then take the value of t and input it into v'(t)

    when solving v(t) = 500 I end up with the answer 10-5sqrt(2) or -5(sqrt(2) - 2)

    but to get the correct answer to the question it needs to be 5sqrt2

    the correct answer is 100sqrt(2) litres/day

    thank you for your help!
    \displaystyle \begin{align*} 500 &= 1000 \left( 1 - \frac{t}{10} \right) ^2 \\ \frac{1}{2} &= \left( 1 - \frac{t}{10} \right) ^2 \\ \pm \sqrt{ \frac{1}{2} } &= 1 - \frac{t}{10} \\ \pm \frac{\sqrt{2}}{2} &= 1 - \frac{t}{10} \\ \frac{t}{10} &= 1 \pm \frac{\sqrt{2}}{2} \\ \frac{t}{10} &= \frac{2 \pm \sqrt{2}}{2} \\ t &= 5 \left( 2 \pm \sqrt{2} \right) \\ t &= 10 \pm 5\sqrt{2} \end{align*}

    Obviously since \displaystyle \begin{align*} 0 \leq t \leq 10 \end{align*} we must accept \displaystyle \begin{align*} t = 10 - 5\sqrt{2} \end{align*}, so I agree with your value of t.


    You should double check your derivative, as it should be \displaystyle \begin{align*} v'(t) = -2000 \left( 1 - \frac{t}{10} \right) \end{align*}, not \displaystyle \begin{align*} -200 \left( 1 - \frac{t}{10} \right) \end{align*}.
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    Re: issues with the algebra in a problem

    ah okay, thank you! I thought the derivative inside the parenthesis was (-1/10) making for

    1000(2)[1-(t/10)](-1/10) -> 2000[1-(1 - (t/10)](-1/10) -> -200[1 - (t/10)]
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    Re: issues with the algebra in a problem

    Oops, you are right, I misread it >_< disregard my last response...
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    Re: issues with the algebra in a problem

    haha no worries. if you could, would you show me how you would work through v'(t) for t = 10-5sqrt2 because I am not getting the correct answer of 100sqrt2

    Thank you!
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    Re: issues with the algebra in a problem

    Well let's see...

    \displaystyle \begin{align*} v' \left( 10 - 5\sqrt{2} \right) &= -200 \left( 1 - \frac{10 - 5\sqrt{2}}{10} \right) \\ &= -200 \left( \frac{10}{10} - \frac{10 - 5\sqrt{2}}{10} \right) \\ &= -200 \left[ \frac{10 - \left( 10 - 5\sqrt{2} \right) }{10} \right] \\ &= -200 \left( \frac{10 - 10 + 5\sqrt{2}}{10} \right) \\ &= -200 \left( \frac{5\sqrt{2}}{10} \right) \\ &= -200 \left( \frac{\sqrt{2}}{2} \right) \\ &= -100\sqrt{2} \end{align*}

    And of course the negative value represents the fact that the water level is decreasing...
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  7. #7
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    Re: issues with the algebra in a problem

    ah I see where I messed up!

    Thank you!!
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